Recent content by Caeder

How do I determine what the original matrix was that yielded these two eigenvalues with the corresponding eigenvectors: \lambda_1 = -3 Eigenvector: [0,1] \lambda_2 = 2 Eigenvector: [1,0] I've played around with det(A-lambda I) but can't find the matrix! I even just did some trial and...

Ha, then I bet for a Real Analysis course this proof would be unacceptable :-p

I have to prove x^3 is differentiable at x=4 using the definition of what it means for something to be differentiable. So I was wondering if I just have to show that f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4} exists, where f(x) = x^3. So... f'(4) = \lim_{x \to 4} \frac {x^3 -...

\begin{array}{l} x_2 < x_3 < 9 \\ \sqrt {x_2 } < \sqrt {x_3 } < 3 \\ 3 + \sqrt {x_2 } < 3 + \sqrt {x_3 } < 6 \\ \underbrace {\sqrt {3 + \sqrt {x_2 } } }_{x_3 } < \underbrace {\sqrt {3 + \sqrt {x_3 } } }_{x_4 } < \sqrt 6 \\ x_3 < x_4 < 9 \\ \end{array} Now I show it from the k+1...

I already know the limit. I'm having a hard time with the proof.

Stable point..

Given the following sequence: x_0 = 1, \quad x_1 = \sqrt{3+1}, \quad x_2 = \sqrt{3+\sqrt{4}}, \quad x_3 = \sqrt{3+\sqrt{5}}, x_4 = \sqrt{3+\sqrt{3+\sqrt{5}}}, \quad x_5 = \sqrt{3+\sqrt{3+\sqrt{3+\sqrt{5}}}} \ldots prove the above sequence converges and determine the limit. ...

We know that the derivative is always negative for values > 1. So we have: f(y) - f(x) \rightarrow \frac{\frac{y}{y-1} - \frac{x}{x-1}}{y-x} = \frac{-1}{(x-1)(y-1)} ... Not sure what this quite does... we know this equals |f'(c)|*|y-x| from what you showed me. f'(c) will have to be 1.5...

Well, looking in my notes, to show f(x) = \frac{1}{x} is unif. continuous on A = [a, \infty) for a > 0, they prove it with: We note x,y \in A, then: |f(x) - f(y)| = \left|\frac{y-x}{xy}\right| \leq \frac{|y-x|}{a^2} Here, if we have that \epsilon > 0 and if |x-y| < a^2\epsilon, then...

Indeed, I meant 1. Typo. I'll look for "Lipschitz" although it doesn't sound familiar.

Define h : \mathbb{R} \rightarrow \mathbb{R} h(x) = \begin{cases} 0 &\text{if\ }\ x \in \mathbb{Q}\\ x^3 + 3x^2 &\text{if\ }\ x \notin \mathbb{Q} \end{cases}. a.) Determine at what points h is continuous and discontinuous. Prove results. b.) Determine at what points h is...

Uh, \frac{-1}{(x-2)^2} ... I don't see how f'(x) helps in showing f(x) is uniformly continuous. We know every function that is integrable is continuous, but we can't say the same with differentiability. Mind helping?

Let f(x) = \frac{x}{x-1}. Prove f(x) is uniformly cont. on the interval [1.5,\infty)